A cylindrical bar of metal having a diameter of 20.9 mm and a length of 203 mm is deformed elastically in tension with a force of 46600 N. Given that the elastic modulus and Poisson's ratio of the metal are 64.4 GPa and 0.33, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

=  0.428 mm

 = −0.0145 mm

Explanation:

given data

diameter = 20.9 mm

length = 203 mm

force = 46600 N

elastic modulus = 64.4 GPa

Poisson's ratio  = 0.33

solution

we get here first area of cross section that is

Area =     .............1

put here value and we will get

Area =

Area = 343.06 mm²    

and

now we get here change in direction of applied stress that is

=     .....................2

put here value and we will get

=    

=  0.428 mm

and

we know Poisson ratio will be here express as

m =       ........................3

0.33 =    

 = −0.0145 mm

so here change in diameter have -ve sign so diameter is decrease.

this is true

explanation:

i looked it up